Six congruent copies of the parabola $y = x^2$ are arranged in the plane so that each vertex is tangent to a circle, and each parabola is tangent to its two neighbors.  Find the radius of the circle.

[asy]
unitsize(1 cm);

real func (real x) {
  return (x^2 + 3/4);
}

path parab = graph(func,-1.5,1.5);

draw(parab);
draw(rotate(60)*(parab));
draw(rotate(120)*(parab));
draw(rotate(180)*(parab));
draw(rotate(240)*(parab));
draw(rotate(300)*(parab));
draw(Circle((0,0),3/4));
[/asy]
Solution: Let $r$ be the radius of the circle.  Then we can assume that the graph of one of the parabolas is $y = x^2 + r.$

Since $\tan 60^\circ = \sqrt{3},$ the parabola $y = x^2 + r$ will be tangent to the line $y = x \sqrt{3}.$

[asy]
unitsize(1 cm);

real func (real x) {
  return (x^2 + 3/4);
}

path parab = graph(func,-1.5,1.5);

draw(dir(240)--3*dir(60),red);
draw(parab);
draw(Circle((0,0),3/4));
draw((-2,0)--(2,0));

label("$60^\circ$", 0.5*dir(30));

dot((0,0),red);
[/asy]

This means the equation $x^2 + r = x \sqrt{3},$ or $x^2 - x \sqrt{3} + r = 0$ will have exactly one solution.  Hence, the discriminant will be 0, so $3 - 4r = 0,$ or $r = \boxed{\frac{3}{4}}.$